## You are here

Homecomposition of multiplicative functions

## Primary tabs

# composition of multiplicative functions

###### Theorem.

If $f$ is a completely multiplicative function and $g$ is a multiplicative function, then $f\circ g$ is a multiplicative function.

###### Proof.

First note that $(f\circ g)(1)=f(g(1))=f(1)=1$ since both $f$ and $g$ are multiplicative.

Let $a$ and $b$ be relatively prime positive integers. Then

$(f\circ g)(ab)$ | $=f(g(ab))$ |

$=f(g(a)\cdot g(b))$ since $g$ is multiplicative | |

$=f(g(a))f(g(b))$ since $f$ is completely multiplicative | |

$=(f\circ g)(a)(f\circ g)(b)$. |

∎

Note that the assumption that $f$ is *completely* multiplicative (as opposed to merely multiplicative) is essential in proving that $f\circ g$ is multiplicative. For instance, $\tau\circ\tau$, where $\tau$ denotes the divisor function, is not multiplicative:

$(\tau\circ\tau)(2\cdot 3)=(\tau\circ\tau)(6)=\tau(\tau(6))=\tau(4)=3$ |

$(\tau\circ\tau)(2)\cdot(\tau\circ\tau)(3)=\tau(\tau(2))\cdot\tau(\tau(3))=\tau% (2)\cdot\tau(2)=2\cdot 2=4$ |

Major Section:

Reference

Type of Math Object:

Theorem

Parent:

## Mathematics Subject Classification

11A25*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections